I guess C_n is the cyclic group of order n. But still, what the OP posted doesn't make a lot of sense, e.g. what's the distinction between F and [itex]\mathbb{F}[/itex]? Also, is [itex]\Re[/itex] actually [itex]\mathbb{R}[/itex]? (The former is the "real part" of a complex number, the latter is the set of real numbers.) And, in the last two, what is C? Finally, in what context are the direct products being taken?

obviously i haven't made myself clear, i'll post the question in it's entirety...

Let A,B be algebras over a field F. We say that A and B are isomorphic over F written [tex] A\cong_F B [/tex] when there exists a bijective ring homomorphism [tex] \varphi : A \rightarrow B [/tex] which is also linear over F, i.e. satisfies [tex] \\\varphi(a\lambda) = \varphi(a) \lambda \mbox{ for all a} \in A \ \lambda \in F [/tex]

here R = field of real numbers, C = field of complex numbers, F is an arbitrary field although in the question it is constrained by the given condition. Also all C_n's are cyclics and by F[C_n] we mean the group ring of C_n over F, that is the ring whose elements are linear combinations of the group elements with coefficients in F.

For some reason this problem popped into my head today, so I gave it another go and managed to solve it. There are two things we need:
(1) [itex]F[C_n] \cong_F F[x]/(x^n - 1)[/itex]. This can be easily proved using the evaluation homomorphism at a generator of C_n.
(2) The http://planetmath.org/encyclopedia/ChineseRemainderTheorem2.html [Broken]. (The CRT is something that slipped my mind at the time.)